We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
-1
194
2
avatar+283 

1. Triangle ABC has side lengths AB=9, AC=10, and BC=17. Let X be the intersection of the angle bisector of angle A with side \(\overline{BC}\), and let Y be the foot of the perpendicular from X to side \(\overline{AC}\). Compute the length of \(\overline{XY} \)

 

2. \(\overline{BC}\) is a chord of a circle with center O and area \(48\pi \). Point A is inside \(\triangle BCO\) such that \(\triangle ABC\) is equilateral and A is the circumcenter of \(\triangle BCO\). What is the area of triangle ABC?

 

 

 

Thank you!!! laugh      

 Apr 6, 2019
edited by Melody  Apr 7, 2019
 #1
avatar+105238 
0

First one....see the image below :

 

 

Let A = (0,0)    

Let C = (10,0)

And we can find B, thusly

 

Construct a circle with radius 9 centered at the origin

The equation of this circle will be  x^2 + y^2 = 81      (1)

 

And construct a circle wilth radius 17 centered at (10, 0)

The equation of this circle will be  (x - 10)^2 + y^2 = 289  (2)

 

We can find the coordinates for point B, thusly :

 

Subtract  (1) from (2)  and we have that

(x-10)^2 - x^2  = 208    simplify

x^2 - 20x + 100 - x^2  = 208

-20x = 108

x = -108/20  = -5.4 =   x coordinate of B

And taking the positive value for y, we have that

 

(-5.4)^2 + y^2 = 81

y^2 = 81 - 29.16

y = 7.2  =  y coordinate of B

 

So....B =  (-5.4, 7.2)

 

And  AX is the bisector of angle A

So....we have the following relationship

AC/AB  = XC /XB

10/9  = XC / XB

Then BC has 19 equal parts and XC is 10 of them.....so  XC = (10/19)*17 = 170/19

 

 

So....using a similar triangle idea.....

 

Distance from B to the x axis / BC  =  XY / XC

 

7.2 / 17  = XY / (170/19)

 

(170/19) * 7.2 /17  = XY  =  (170/17)(7.2 / 19) = 10 * 7.2 / 19    =   72 / 19

 

 

cool cool cool

 Apr 6, 2019
 #2
avatar+105238 
0

Second one.....see the following image....

 

 

Let O = (0,0)

Since the area of circle O = 48pi....then the radius  = √48  units

So...the equation of the circle is x^2 + y^2 = 48

Since ABC is equilateral....then angle BAC = 60°

Then the excluded angle to  BAC = 300°

So, by symmetry, angle  AOC = 150°

And since A is the circumcenter of triangle BOC, then AO = AC = the side of equilateral triangle ABC = s

 

And we can find s^2 using the Law of Cosines, thusly :

 

OC^2 = 2s^2 - 2s^2 * ( cos 150)

48 = 2s^2 + 2s^2* (√3/2)

48 = 2s^2 + s^2√3

48 = (2 + √3)s^2

48 / [ 2 + √3 )  = s^2    = 48 [ 2 - √3]  =  96 - 48√3

 

So....the area of equilateral triangle ABC  = 

 

(1/2) s^2 sin (BAC)  = 

 

(1/2)s^2 sin(60°)  =

 

(√3) [ 96 - 48√3 ] / 4  =

 

(√3) [ 24 - 12√3]  =   

 

24√3 - 36   units^2

 

 

cool cool cool

 Apr 7, 2019

12 Online Users